$\begin{array}{l} \left\{ \begin{array}{l} \left( {x + 2y + 1} \right)\left( {x + 2y + 2} \right) = 0\\ xy + {y^2} + 3y + 1 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x = - 2y - 1\left( 1 \right)\\ x = - 2y - 2\left( 2 \right) \end{array} \right.\\ xy + {y^2} + 3y + 1 = 0\left( 3 \right) \end{array} \right.\\ \left( 1 \right) \to \left( 3 \right):\left( { - 2y - 1} \right)y + {y^2} + 3y + 1 = 0\\ \Leftrightarrow - {y^2} + 2y + 1 = 0\\ \Leftrightarrow {y^2} - 2y - 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = 1 + \sqrt 2 \Rightarrow x = - 3 - 2\sqrt 2 \\ y = 1 - \sqrt 2 \Rightarrow x = - 4 + 2\sqrt 2 \end{array} \right.\\ \left( 2 \right) \to \left( 3 \right):\left( { - 2y - 2} \right)y + {y^2} + 3y + 1 = 0\\ \Leftrightarrow - {y^2} + y + 1 = 0\\ \Leftrightarrow {y^2} - y - 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt 5 }}{2} \Rightarrow y = - 3 - \sqrt 5 \\ x = \dfrac{{1 - \sqrt 5 }}{2} \Rightarrow y = - 3 + \sqrt 5 \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( { - 3 - 2\sqrt 2 ;1 + \sqrt 2 } \right),\left( { - 4 + 2\sqrt 2 ;1 - \sqrt 2 } \right),\left( {\dfrac{{1 + \sqrt 5 }}{2}; - 3 - \sqrt 5 } \right),\left( {\dfrac{{1 - \sqrt 5 }}{2}; - 3 + \sqrt 5 } \right)\\ \left\{ \begin{array}{l} \left( {x + y + 2} \right)\left( {2x + 2y - 1} \right) = 0\\ 3{x^2} + 32{y^2} + 5 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x = - y - 2\left( 1 \right)\\ x = \dfrac{{1 - 2y}}{2}\left( 2 \right) \end{array} \right.\\ 3{x^2} + 32{y^2} + 5 = 0\left( 3 \right) \end{array} \right.\\ \left( 1 \right) \to \left( 3 \right):3{\left( { - y - 2} \right)^2} + 32{y^2} + 5 = 0\\ \Leftrightarrow 9{y^2} + 12y + 12 + 32{y^2} + 5 = 0\\ \Leftrightarrow 41{y^2} + 12y + 17 = 0\left( {PTVN} \right)\\ \left( 2 \right) \to \left( 3 \right):3.{\left( {\dfrac{{1 - 2y}}{2}} \right)^2} + 32{y^2} + 5 = 0\\ \Leftrightarrow \dfrac{{3{{\left( {2y - 1} \right)}^2}}}{4} + 32{y^2} + 5 = 0\\ \Leftrightarrow 3{\left( {2y - 1} \right)^2} + 128{y^2} + 20 = 0\\ \Leftrightarrow 140{y^2} - 12y + 23 = 0\\ \Rightarrow PTVN\\ \left( {x;y} \right) = \emptyset \end{array}$
